# Derivatives

## The derivative of ${\frac {d}{dx}}x^{x}$ ### Question

What is ${\frac {d}{dx}}x^{x}$ ?

### Solution 1

• Let $y=x^{x}$ .
• Take $\ln$ of both sides: $\ln y=x\ln x$ .
• Differentiate both sides: ${\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln x$ .
• Apply the chain rule on the left-hand side: ${\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {dy}{dx}}$ .
• Apply the product rule on the right-hand side: ${\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1$ .
• Putting it together, we have ${\frac {1}{y}}\cdot {\frac {dy}{dx}}=\ln x+1$ .
• Hence ${\frac {dy}{dx}}=y(\ln x+1)=x^{x}(\ln x+1)$ .

### Solution 2

• Note that $x=e^{\ln x}$ , so $x^{x}=(e^{\ln x})^{x}=e^{x\ln x}$ .
• Applying the chain rule, ${\frac {d}{dx}}x^{x}={\frac {d}{dx}}e^{x\ln x}=e^{x\ln x}{\frac {d}{dx}}x\ln x$ .
• Applying the product rule, ${\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1$ .
• Therefore ${\frac {d}{dx}}x^{x}=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)$ .

# Integrals

## The integral $\int x^{x}\,dx$ ### Question

What is $\int x^{x}\,dx$ ?

### Solution

• We can write $x^{x}$ as $(e^{\ln x})^{x}=e^{x\ln x}$ .
• Consider the series expansion of $e^{x\ln x}$ :
   $e^{x\ln x}=1+(x\ln x)+{\frac {(x\ln x)^{2}}{2!}}+{\frac {(x\ln x)^{3}}{3!}}+\ldots +{\frac {(x\ln x)^{i}}{i!}}+\ldots =\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}$ .

• We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write
   $\int x^{x}\,dx=\int \left(\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}\right)\,dx=\sum _{i=0}^{\infty }\left(\int {\frac {(x\ln x)^{i}}{i!}}\,dx\right)=\sum _{i=0}^{\infty }\left({\frac {1}{i!}}\int x^{i}(\ln x)^{i}\,dx\right).$ 